The freezing point of aqueous solution that contains 5% by mass urea, 1.0% by mass KCl and 10% by mass of glucose is ( $K_{f (H_{2} O)} = 1.86 K$ Molality $^{- 1}$ ) :

290.2 K

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285.5 K

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269.48 K

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250 K

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Solution

The correct option is C 269.48 K
The aqueous solution contains 5% by mass urea, 1.0% by mass KCl and 10% by mass of glucose.
Suppose we have 100 g of solution. It will contain 5 g urea, 1.0 g KCl and 10 g glucose.
The molar masses of urea, KCl and glucose are 60 g/mol, 74.5 g/mol and 180 g/mol respectively.
The number of moles of urea $= \frac{5 g}{60 g / m o l}$
The number of moles of KCl $= \frac{1 g}{74.5 g / m o l}$
The number of moles of glucose $= \frac{10 g}{180 g / m o l}$
The depression in the freezing point depends on the number of solute particles. Glucose and urea are non electrolytes. So they do not decompose.
KCl is a strong electrolyte. It dissociates in aqueous solution in K+ ions and Cl- ions. So number of moles of KCl should be multiplied by 2.
Mass of solvent is $100 - [5 + 1 + 10] = 84 g$
The molality of the solution is the ratio of the total number of moles of solute particles to mass of solvent in kg.
Molality $m = \frac{\frac{5 g}{60 g / m o l} + 2 \times \frac{1 g}{74.5 g / m o l} + \frac{10 g}{180 g / m o l}}{84 g} \times 1000 g / k g = 1.97 m$
The depression in the freezing point $Δ T_{f} = K_{f} m = 1.86 \times 1.97 = 3.66$
The freezing point of water is 273.15 K.
The freezing point of aqueous solution is
$273.15 - 3.66 = 269.48$ K

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The freezing point of aqueous solution that contains 5% by mass urea, 1.0% by mass KCl and 10% by mass of glucose is (Kf(H2O)=1.86K Molality−1) :

The freezing point of aqueous solution that contains 5% by mass urea, 1.0% by mass KCl and 10% by mass of glucose is ( $K_{f (H_{2} O)} = 1.86 K$ Molality $^{- 1}$ ) :